lacy has $600.00 set aside to build a rectangular excercise kennel for her dogs. She will buy fencing material for $15/ft. Because the side of an existing barn will be used for one of the sides of the kennel, only three sides need to be fenced. Lacy needs the area of the kennel to be 150^2. Determine the dimensions of a kennel that uses all possible fencing material. Using vertex form find the dimensions .
6 years ago
Answered By Negar S
Lacy has $600 and fencing material costs $15/ft, that means Lacy can build how many feet of fence ?
L=600/15 = 40ft
she wants to fence an area of 150 $\frac{-\left(-40\right)+\sqrt{\left(-40\right)^2-4\times2\times150}}{2\times2}=\frac{40+\sqrt{400}}{4}=15ft$−(−40)+√(−40)2−4×2×1502×2 =40+√4004 =15ƒ t , I assume we are talking about ft2
that means if she has a rectangular area, we name one side of rectangular a and the other side b :
a.b =150 (ft2)
and total perimeter that Lacy wants to fence is :
L= a + b +a
this is equal to 40 ft
now we have two equations :
2a+b=40 (1)
a.b=150 (2)
from equation (2), b=150/a
equation (1) : 2a + 150/a =40
I can write equation (1) :
2a2-40a+150=0
do you remember quadratic formula ? :
a= $\frac{-\left(-40\right)+\sqrt{\left(-40\right)^2-4\times2\times150}}{2\times2}=\frac{40+\sqrt{400}}{4}=15ft$−(−40)+√(−40)2−4×2×1502×2 =40+√4004 =15ƒ t
b=150/a b=150/15=10ft
or
a= $\frac{-\left(-40\right)-\sqrt{\left(-40\right)^2-4\times2\times150}}{2\times2}=\frac{40-\sqrt{400}}{4}=5ft$−(−40)−√(−40)2−4×2×1502×2 =40−√4004 =5ƒ t
b=150/5 = 30 ft
Now vertex form :
0=2(a-h)2+k
2h2+k=150
-4h=-40
h=10
k=150-200= -50
2(a-10)2-50=0
(a-10)2=25
a-10= 5
a=15ft
or
a-10=-5
a=15ft
both answers are matching so Lacy can make a fence with sides of either 5ft by 30ft or 15ft by10ft.